I need to prove: If $p$ is a prime number congruent to $5 \mod 8$, and $\left(\frac np\right)= 1$, then

either $ [n^{(p+3)/8}] ^2 ≡ n\bmod p$

or $ [n^{(p+3)/8}((p-1)/2)! ]^2 ≡ n\bmod p$

I am not really sure how to proceed, although I have tried checking the cases $n= 1$ and $n = -1$, I have also tried various things with the definition of a Legendre symbol.


You can argue as follows.

We know that $n$ is a quadratic residue. The question is whether $n$ is a quartic residue or not.

  • Assume first that $n\equiv a^4\pmod p$ for some $a$ coprime to $p$. By Little Fermat we can deduce that $$1\equiv a^{p-1}\equiv n^{(p-1)/4}\pmod p.$$ Multiplying this congruence by $n$ gives $$ n^{(p+3)/4}\equiv n\pmod p. $$ As $k=(p+3)/8$ is an integer, this means that $(n^k)^2\equiv n$ in line with the first alternative.
  • If $n$ is a modular square but not a modular fourth power, then I first claim that $-n$ will be a fourth power modulo $p$. This is because $-1$ is known to be a quadratic residue. But as $p\not\equiv 1\pmod 8$ there are no elements of order $8$, and consequently $-1$ cannot be a fourth power. The claim follows from this because the quotient group of quadratic residues modulo quartic residues is cyclic of order two (this holds for all $p\equiv1\pmod4$). The result of the previous bullet thus implies the congruence $$ (-n)^{(p+3)/4}\equiv -n\pmod p. $$ Here $(p+3)/4$ is even, so this reads $$ n^{(p+3)/4}\equiv -n\pmod p.\qquad(*) $$
  • Let $u$ be the residue class of $((p-1)/2)!$. Because $p\equiv1\pmod4$ it follows from Wilson's theorem, $(p-1)!\equiv-1\pmod p$, that $u^2\equiv-1\pmod p$. This allows us to rewrite congruence $(*)$ in the form $$[u n^{(p+3)/8}]^2\equiv n^{(p+3)/4}u^2\equiv (-n)\cdot (-1)=n\pmod p,$$ which is exactly what the second alternative says.
  • $\begingroup$ yes, you are correct about the typo. I am not good with MathJax on here. $\endgroup$ – user2782067 Nov 1 '18 at 19:11
  • $\begingroup$ Thanks for notifying me about the fix @user2782067. Don't worry. You'll get the hang of it quickly enough. $\endgroup$ – Jyrki Lahtonen Nov 1 '18 at 19:13

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